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Bài 1:
a)12,5 x (-5/7) + 1,5 x (-5/7)
=-5/7*(12,5+1,5)
=-5/7*14
=-10
b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)
=-1/4*(68/11+42/11)
=-1/4*10
=-5/2
c tương tự
d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)
Bài 2:
a)Ta có:
2800=(28)100=256100
8200=(82)100=64100
Vì 256100>64100 =>2800>8200
b)Ta có:
1245=(123)15=172815
Vì 62515<172815 =>62515<1245
a,12,5x(-5/7)+1,5x(-5/7)
=-125/14+-15/14
=-10
2,2mu800>8 mu 200
6254 lon hon 12
\(11^{12}< 11^{13}\)
\(7^4< 8^4\)
\(3^4>4^3\)
\(2^6>6^2\)
\(5^{15}>2^{30}\)
1. A - B = 40+ 3/8 + 7/82 + 5/83 + 32/85 - (24/82 + 40+ 5/82 + 40/84 + 5/84 )
= 40.85/85 + 3.84/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 40.85/85 + 24.83/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 + 32/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 -8/85 - 5.83/85 - 40.8/85
= 2.83/85 + 5.82/85 - 40.8/85 - 8/85
= 2.83/85 + 40.8/85 - 40.8/85 - 8/85
= 2.83/85 - 8/85 > 0
Vay A > B
\(A=3^2-3^5+3^8-3^{11}+...-3^{101}\)
\(\Rightarrow3A=3^5-3^8+3^{11}-3^{14}+...-3^{104}\)
\(\Rightarrow3A+A=\left(3^5-3^8+3^{11}-3^{14}+...-3^{104}\right)+\left(3^2-3^5+3^8-3^{11}+...-3^{101}\right)\)
\(\Rightarrow4A=-3^{104}+3^2\)
\(\Rightarrow28A=7\left(3^2-3^{104}\right)\)
\(\Rightarrow B+28A=3^{104}+7\left(3^2-3^{104}\right)\)
\(\Rightarrow B+28A=7.3^2-6.3^{104}=3^2\left(7-2.3^{103}\right)\)
B1:
Ta có: \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}-2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.\left(-4\right)}{3.5}=-\frac{8}{15}\)
B2:
Ta có: \(1+3+5+...+x=1600\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot\left(\frac{x-1}{2}+1\right)}{2}=1600\)
\(\Leftrightarrow\left(x+1\right)\cdot\frac{x+1}{2}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=6400\)
Xét theo dãy tăng tiến ta thấy được giá trị của x càng tăng
=> x dương => x + 1 dương
\(\Rightarrow x+1=80\)
\(\Rightarrow x=79\)
a) \(21^{15}\)\(=\left(3.7\right)^{15}\) \(=3^{15}.7^{15}\)
\(27^5.49^8\) \(=\left(3^3\right)^5.\left(7^2\right)^8\)\(=3^{15}.7^{16}\)
Ta thấy \(7^{15}< 7^{16}\)\(\Rightarrow\)\(21^{15}< 27^5.49^8\)
b) \(3^{30}\)\(=3^{2.15}\)\(=\left(3^2\right)^{15}\)\(=9^{15}\)
Ta thấy \(8^{15}< 9^{15}\)
\(\Rightarrow\)\(8^{15}< 3^{30}\)
Ta có: \(21^{15}=\left(3.7\right)^{15}=3^{15}.7^{15}\)
\(27^5.49^8=3^{3.5}.7^{2.8}=3^{15}.7^{16}\)
Vì \(15< 16\)nên \(7^{15}< 7^{16}\Rightarrow3^{15}.7^{15}< 3^{15}.7^{16}\)
Hay \(21^{15}< 27^5.49^8\)
Ta có: \(3^{30}=3^{2.15}=9^{15}\)
Vì \(8< 9\)nên \(8^{15}< 9^{15}\)
Hay \(8^{15}< 3^{30}\)
Bài làm
a) Ta có: 26 = ( 23 )2 = 82
Mà 8 > 6
=> 82 > 62
Hay 26 > 62
Vậy 26 > 62
# Học tốt #
Bài làm
c) Ta có: 215 = ( 23 )5 = 85
Mà 8 = 8
=> 85 = 85
Hay 215 = 85
Vậy 215 = 85
# Học tốt #